$h(t) = -5t^{3}-2t^{2}+2+3(f(t))$ $f(t) = -t-4$ $g(x) = -4x^{2}-x+3(f(x))$ $ f(g(-2)) = {?} $
First, let's solve for the value of the inner function, $g(-2)$ . Then we'll know what to plug into the outer function. $g(-2) = -4(-2)^{2}-(-2)+3(f(-2))$ To solve for the value of $g$ , we need to solve for the value of $f(-2)$ $f(-2) = -(-2)-4$ $f(-2) = -2$ That means $g(-2) = -4(-2)^{2}-(-2)+(3)(-2)$ $g(-2) = -20$ Now we know that $g(-2) = -20$ . Let's solve for $f(g(-2))$ , which is $f(-20)$ $f(-20) = -(-20)-4$ $f(-20) = 16$